P4.1 Proof by contradiction and partial fractions
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
A2 content for Pure Mathematics 4 (specification sections 1.1 and 2.1). Denominators with a quadratic factor that will not factorise are not needed.
What this page covers
7 points
- The structure of a proof by contradiction, and how to write the negation of a statement correctly
- The proof that \(\sqrt2\) is irrational, including the step "if \(a^2\) is even then \(a\) is even"
- The proof that there are infinitely many prime numbers
- Proof by contradiction for unfamiliar statements: rational and irrational numbers, equations with no integer solutions, "no greatest" results
- Partial fractions with distinct linear factors in the denominator (two or three factors)
- Partial fractions with a repeated linear factor
- Improper fractions: divide first, then split
In one minute
5 points
- A proof by contradiction assumes the statement is false, reasons correctly from that assumption to something impossible, and concludes that the statement must be true; partial fractions do the reverse of adding fractions, writing one complicated fraction as a sum of simple ones.
- \(\sqrt2\) is irrational: assume \(\sqrt2 = \frac{a}{b}\) in its simplest form; squaring shows \(a\) and \(b\) are both even, a common factor, which contradicts "simplest form".
- Infinitely many primes: assume a complete finite list; the product of the list plus 1 is not divisible by any prime in the list, so a prime is missing.
- Distinct linear factors: \(\frac{f(x)}{(x-p)(x-q)} \equiv \frac{A}{x-p} + \frac{B}{x-q}\); a repeated factor needs both \(\frac{B}{x-p}\) and \(\frac{C}{(x-p)^2}\).
- Improper fraction (degree of numerator ≥ degree of denominator): divide first, then split the proper remainder fraction.
Contents
14 sections
- How does a proof by contradiction work?
- How do you prove that √2 is irrational?
- How do you prove there are infinitely many primes?
- How do you tackle an unfamiliar statement?
- How do you split a fraction with distinct linear factors?
- What changes when a factor is repeated?
- What do you do with an improper fraction?
- How do the two ideas combine in a longer question?
- Exam tips for P4.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How does a proof by contradiction work?
This page contains the last proof method of the course and the algebra tool that the rest of P4 leans on. In P2 you proved statements directly (deduction), by checking every case (exhaustion), and disproved them with a counterexample. Some statements resist all three, usually because they say something cannot happen: a number is not a fraction, a list never ends, an equation has no solution. For these, it is easier to imagine that the statement fails and show that this imagined world is impossible.
Stage 1 is where most proofs go wrong, so it pays to practise writing the negation of a statement.
| Statement to prove | Assume, for contradiction, that … |
|---|---|
| \(\sqrt2\) is irrational | \(\sqrt2\) is rational: \(\sqrt2 = \frac{a}{b}\) with \(a, b\) integers with no common factor, \(b \ne 0\) |
| There are infinitely many primes | there are finitely many primes, \(p_1, p_2, \ldots, p_n\) (a complete list) |
| There is no greatest even integer | there is a greatest even integer, \(N\) |
| If \(n^2\) is even, then \(n\) is even | \(n^2\) is even and \(n\) is odd |
| There are no integers \(x, y\) with \(x^2 - y^2 = 10\) | there are integers \(x, y\) with \(x^2 - y^2 = 10\) |
The negation of "every … has property P" is "at least one … does not have property P", and the negation of "there is no …" is "there is at least one …". The negation of "if P then Q" is "P is true and Q is false": keep the condition, deny the result.
A quick example: there is no greatest even integer. Assume \(N\) is the greatest even integer. Then \(N + 2\) is even and \(N + 2 > N\), which contradicts \(N\) being the greatest. So there is no greatest even integer.
Negating "if \(n^2\) is even then \(n\) is even" as "if \(n^2\) is odd then \(n\) is odd". That is a different statement altogether. The correct starting point keeps the condition and denies the conclusion: "\(n^2\) is even and \(n\) is odd".
How do you prove that √2 is irrational?
You need one small result first: if \(a^2\) is even, then \(a\) is even. It is itself a proof by contradiction. Assume \(a^2\) is even and \(a\) is odd, so \(a = 2k + 1\) for some integer \(k\). Then \(a^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\), which is odd, contradicting "\(a^2\) is even". So if \(a^2\) is even, \(a\) is even. Quote this result (or prove it) inside the \(\sqrt2\) proof.
Prove by contradiction that \(\sqrt2\) is irrational.
Assume the opposite. Assume \(\sqrt2\) is rational: \(\sqrt2 = \frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \ne 0\), and \(a\) and \(b\) have no common factor other than 1.
Square and rearrange. \(2 = \frac{a^2}{b^2} \Rightarrow a^2 = 2b^2\), so \(a^2\) is even, and therefore \(a\) is even.
Write \(a = 2c\) and substitute. \((2c)^2 = 2b^2 \Rightarrow 4c^2 = 2b^2 \Rightarrow b^2 = 2c^2\), so \(b^2\) is even and \(b\) is even.
Contradiction. \(a\) and \(b\) are both even, so they have a common factor 2. This contradicts the assumption that \(\frac{a}{b}\) is in its simplest form.
Conclude. The assumption is false, so \(\sqrt2\) is irrational.
Every rational number can be written in simplest form, so assuming this costs nothing, and it is exactly what the contradiction breaks.
Replace "even" by "a multiple of 3" (or 5, or 7). You then need the matching result, if \(a^2\) is a multiple of 3 then \(a\) is a multiple of 3, which you can prove by considering \(a = 3k + 1\) and \(a = 3k + 2\). Do not try it for \(\sqrt4\): the argument breaks, because 4 is a square.
How do you prove there are infinitely many primes?
This proof is over two thousand years old and is a model of the method: assume the list of primes is finite, then build a number the list cannot account for.
Prove by contradiction that there are infinitely many prime numbers.
Assume the opposite. Assume there are only finitely many primes, and that \(p_1, p_2, \ldots, p_n\) is the complete list of them.
Build a new number. Let \(N = p_1 p_2 \cdots p_n + 1\). Then \(N > 1\) and \(N\) is greater than every prime in the list.
Test \(N\) against the list. Dividing \(N\) by any \(p_i\) leaves a remainder of 1, so no prime in the list is a factor of \(N\).
Contradiction. Every integer greater than 1 is either prime or has a prime factor. If \(N\) is prime, it is a prime not in the list; if not, its prime factors are not in the list. Either way the list was not complete.
Conclude. The assumption is false, so there are infinitely many primes.
Writing "\(N = p_1 p_2 \cdots p_n + 1\) is prime, so it is a new prime". \(N\) need not be prime (30 031 is not). The correct argument is that \(N\) has some prime factor, and that factor cannot be in the list.
How do you tackle an unfamiliar statement?
- Write the negation precisely, with every letter defined: "Assume there exist integers \(x\) and \(y\) such that …", "Assume \(a + b\) is rational, so \(a + b = \frac{r}{s}\) …".
- Choose the tool the statement suggests: even/odd (parity), multiples and remainders, the definition of a rational number, or comparing sizes.
- Stop at a clear impossibility: an odd number equal to an even one, a number both rational and irrational, a largest number with something larger.
- Conclude in words that refer back to the original statement.
Given that \(a\) is rational and \(b\) is irrational, prove by contradiction that \(a + b\) is irrational. Hence explain why \(3 + \sqrt2\) is irrational.
Assume \(a + b\) is rational: \(a + b = \frac{r}{s}\), and \(a = \frac{p}{q}\), with \(p, q, r, s\) integers and \(q, s \ne 0\).
Make \(b\) the subject: \(b = \frac{r}{s} - \frac{p}{q} = \frac{rq - ps}{sq}\). Here \(rq - ps\) and \(sq\) are integers and \(sq \ne 0\), so \(b\) is rational, contradicting "\(b\) is irrational".
So \(a + b\) is irrational. With \(a = 3\) (rational) and \(b = \sqrt2\) (irrational, worked example 1), \(3 + \sqrt2\) is irrational.
Prove by contradiction that there are no integers \(x\) and \(y\) such that \(x^2 - y^2 = 10\).
Assume integers \(x, y\) exist with \(x^2 - y^2 = 10\); then \((x - y)(x + y) = 10\).
Compare the two factors: \((x + y) - (x - y) = 2y\), which is even, so \(x - y\) and \(x + y\) are both odd or both even.
Both odd: the product of two odd numbers is odd, but 10 is even. Impossible.
Both even: write them as \(2m\) and \(2n\); the product is \(4mn\), a multiple of 4, but \(10 = 4 \times 2 + 2\) is not. Impossible.
Every case gives a contradiction, so there are no integers \(x, y\) with \(x^2 - y^2 = 10\). The same argument shows that no number of the form \(4k + 2\) is a difference of two squares.
When the statement is about integers, look at parity first. Factorising (a difference of two squares, a common factor) and asking "odd or even?" about each factor finishes many unfamiliar proofs in a few lines.
How do you split a fraction with distinct linear factors?
You know how to add \(\frac{2}{x-1} + \frac{3}{2x+1}\) into a single fraction. Partial fractions run that process backwards. The reason to bother: a single fraction such as \(\frac{7x-1}{(x-1)(2x+1)}\) is hard to integrate or expand, while each simple piece is easy.
Finding the constants. Multiply both sides by the whole denominator to get an identity between numerators, for example \(7x - 1 \equiv A(2x+1) + B(x-1)\). Then:
- Substitution: an identity is true for every \(x\), so choose the values that make a bracket zero; each one isolates a single constant.
- Comparing coefficients: equate the coefficients of \(x^2\), \(x\) and the constant terms on both sides; useful when no convenient substitution is left.
- Factorise the denominator first if it is given expanded, e.g. \(2x^2 + 5x + 2 = (2x+1)(x+2)\).
Express in partial fractions (a) \(\frac{7x-1}{(x-1)(2x+1)}\) (b) \(\frac{x^2 + 11x - 6}{x(x-2)(x+3)}\).
(a) \(7x - 1 \equiv A(2x+1) + B(x-1)\). Substitute the roots of each factor: \(x = 1\): \(6 = 3A \Rightarrow A = 2\); \(x = -\frac12\): \(-\frac92 = -\frac32 B \Rightarrow B = 3\). \(\frac{2}{x-1} + \frac{3}{2x+1}\)
(b) Three factors, three constants: \(x^2 + 11x - 6 \equiv A(x-2)(x+3) + Bx(x+3) + Cx(x-2)\). \(x = 0\): \(-6 = -6A \Rightarrow A = 1\); \(x = 2\): \(20 = 10B \Rightarrow B = 2\); \(x = -3\): \(-30 = 15C \Rightarrow C = -2\). \(\frac1x + \frac{2}{x-2} - \frac{2}{x+3}\) Check by comparing \(x^2\) coefficients: \(A + B + C = 1 + 2 - 2 = 1\).
Substituting the wrong value for a factor such as \((2x+1)\). The bracket is zero when \(x = -\frac12\), not when \(x = -1\) or \(-2\). Fractional values are normal here; work them through carefully.
What changes when a factor is repeated?
Multiplying through by \((ax+b)^2(cx+d)\) gives \(f(x) \equiv A(ax+b)(cx+d) + B(cx+d) + C(ax+b)^2\). Substituting the two roots finds \(B\) and \(C\), but there is no root left to isolate \(A\). Find \(A\) by comparing the coefficients of \(x^2\) (quickest) or by substituting any other value, such as \(x = 0\). Use one further value as a check: it costs a line and catches most slips.
Express \(\frac{3x^2 + 6}{(x-1)^2(x+2)}\) in partial fractions.
Form: \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}\), so \(3x^2 + 6 \equiv A(x-1)(x+2) + B(x+2) + C(x-1)^2\).
Substitute the roots: \(x = 1\): \(9 = 3B \Rightarrow B = 3\); \(x = -2\): \(18 = 9C \Rightarrow C = 2\). Compare coefficients of \(x^2\): \(3 = A + C \Rightarrow A = 1\).
Check with \(x = 0\): \(6 = -2A + 2B + C = -2 + 6 + 2 = 6\). \(\frac{1}{x-1} + \frac{3}{(x-1)^2} + \frac{2}{x+2}\)
Leaving out the \(\frac{A}{x-1}\) term and writing only \(\frac{B}{(x-1)^2} + \frac{C}{x+2}\). Two constants cannot match a numerator with three coefficients, so the equations become inconsistent or give a wrong answer.
What do you do with an improper fraction?
Divide first: numerator \(\equiv\) (denominator) \(\times Q(x) + R(x)\), so the fraction \(= Q(x) + \frac{R(x)}{\text{denominator}}\), with the degree of \(R\) less than the degree of the denominator. Then split the proper fraction \(\frac{R(x)}{\text{denominator}}\) as before. With the same degree on top and bottom the quotient is a constant, e.g. \(\frac{x^2+1}{(x-1)(x+1)} = 1 + \frac{2}{x^2-1} = 1 + \frac{1}{x-1} - \frac{1}{x+1}\). Alternatively, write the full form at once (e.g. \(Ax + B + \frac{C}{\ldots} + \frac{D}{\ldots}\)), multiply through and compare coefficients.
Express \(\frac{2x^3 + 3x^2 - 2x + 6}{(x-1)(x+2)}\) in the form \(Ax + B + \frac{C}{x-1} + \frac{D}{x+2}\).
Divide. \((x-1)(x+2) = x^2 + x - 2\). \(2x^3 \div x^2 = 2x\); subtract \(2x(x^2 + x - 2)\) to leave \(x^2 + 2x + 6\). \(x^2 \div x^2 = 1\); subtract \((x^2 + x - 2)\) to leave \(x + 8\). Quotient \(2x + 1\), remainder \(x + 8\), so the fraction \(= 2x + 1 + \frac{x+8}{(x-1)(x+2)}\).
Split the proper part. \(x + 8 \equiv C(x+2) + D(x-1)\): \(x = 1\): \(9 = 3C \Rightarrow C = 3\); \(x = -2\): \(6 = -3D \Rightarrow D = -2\).
Assemble. \(A = 2\), \(B = 1\), \(C = 3\), \(D = -2\): \(2x + 1 + \frac{3}{x-1} - \frac{2}{x+2}\)
If the top has degree 3 and the bottom degree 2, the quotient is linear (\(Ax + B\)); if the degrees are equal, the quotient is a constant. Forgetting the quotient makes every later constant wrong.
How do the two ideas combine in a longer question?
Once a fraction is in partial fractions, each term is simple: its sign, its derivative and its behaviour for large \(x\) can be read off directly. That makes proofs about the function short, including proofs by contradiction.
\(f(x) = \frac{x^3 + 8x^2 + 20x + 19}{(x+1)^2(x+3)}\), \(x > -1\). (a) Express \(f(x)\) in the form \(A + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{D}{x+3}\). (b) Hence show that \(f\) is decreasing for \(x > -1\). (c) Prove by contradiction that there is no value of \(x > -1\) for which \(f(x) \le 1\).
(a) Degrees are equal, so divide: \((x+1)^2(x+3) = x^3 + 5x^2 + 7x + 3\), so \(A = 1\) and the remainder is \(3x^2 + 13x + 16\). Then \(3x^2 + 13x + 16 \equiv B(x+1)(x+3) + C(x+3) + D(x+1)^2\). \(x = -1\): \(6 = 2C \Rightarrow C = 3\); \(x = -3\): \(4 = 4D \Rightarrow D = 1\); \(x^2\): \(3 = B + D \Rightarrow B = 2\). Check \(x = 0\): \(3B + 3C + D = 6 + 9 + 1 = 16\). \(f(x) = 1 + \frac{2}{x+1} + \frac{3}{(x+1)^2} + \frac{1}{x+3}\)
(b) Differentiate term by term: \(f'(x) = -\frac{2}{(x+1)^2} - \frac{6}{(x+1)^3} - \frac{1}{(x+3)^2}\). For \(x > -1\), \(x + 1 > 0\) and \(x + 3 > 0\), so every term is negative. \(f'(x) < 0\), so \(f\) is decreasing.
(c) Assume some \(x > -1\) has \(f(x) \le 1\); then \(\frac{2}{x+1} + \frac{3}{(x+1)^2} + \frac{1}{x+3} \le 0\). But for \(x > -1\) each of these terms is positive, so their sum is positive: a contradiction. So \(f(x) > 1\) for every \(x > -1\).
Exam tips for P4.1
Open with "Assume, for contradiction, that …" and close with "This is a contradiction, so …". The first line shows which method you are using; the last line shows you know what the contradiction proves.
"\(\sqrt2 = \frac{a}{b}\)" is not enough. Say that \(a\) and \(b\) are integers, \(b \ne 0\), and the fraction is in its simplest form: the contradiction depends on that last condition.
Write the numerator identity in full before substituting. It shows the method even if an arithmetic slip follows, and it makes comparing coefficients easy when a constant cannot be found by substitution. Then check with a spare value, often \(x = 0\).
If the question asks for \(A + \frac{B}{\ldots} + \frac{C}{\ldots}\), give all the constants and the assembled expression. Partial fractions are usually the first part of a longer question, and the later parts use exactly that form.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| To prove "if P then Q" by contradiction, assume "if not P then not Q". | That is a different statement; it does not deny what you want to prove. | Assume P is true and Q is false. |
| In the \(\sqrt2\) proof, "simplest form" is an optional extra. | Without it, finding that \(a\) and \(b\) are both even contradicts nothing. | State the simplest-form condition in the assumption. |
| \(p_1 p_2 \cdots p_n + 1\) is always prime. | \(2 \times 3 \times 5 \times 7 \times 11 \times 13 + 1 = 30\,031 = 59 \times 509\). | It has a prime factor that is not in the list. |
| \(\frac{f(x)}{(x-1)^2(x+2)}\) needs only \(\frac{B}{(x-1)^2} + \frac{C}{x+2}\). | A repeated factor contributes two terms. | Include \(\frac{A}{x-1}\) as well. |
| Any fraction can be split straight away. | For an improper fraction the forms do not match the degrees. | Divide first when the degree of the numerator ≥ the degree of the denominator. |
| Substitution only works for values that appear in the question. | The numerator equation is an identity, true for all \(x\). | Any value may be substituted; roots of the factors are simply the most efficient. |
How this topic connects
P2 Proof and algebraic division
Proof by deduction, exhaustion and counterexample complete the set of proof methods with contradiction; algebraic long division is the first step for every improper fraction.
P3 Algebraic fractions and differentiation
Simplifying and adding algebraic fractions is the reverse of partial fractions, and differentiating each term (as in worked example 8) uses the chain rule from P3. The factors of a denominator give the vertical asymptotes of a curve, as for transformations of \(\frac1x\).
P4.3 The binomial series and P4.6 Integration
Partial fractions turn a fraction into terms that integrate to logarithms, and into terms that can each be expanded with the binomial series. In IAL Physics, resistors in parallel combine as a sum of reciprocals, \(\frac1R = \frac{1}{R_1} + \frac{1}{R_2}\): adding fractions, the reverse of splitting them.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{f(x)}{(ax+b)(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{cx+d}\) | partial fractions, distinct linear factors (proper fraction) | No — learn it |
| \(\frac{f(x)}{(ax+b)^2(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{cx+d}\) | partial fractions with a repeated linear factor | No — learn it |
| \(\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}\), \(\deg R < \deg D\) | improper fractions: divide before splitting | No — learn it |
| \(f(x) \equiv A(cx+d) + B(ax+b)\) | numerator identity used to find the constants | No — derive it |
| \(a\) odd \(\Rightarrow a^2 = 2(2k^2 + 2k) + 1\) odd; so \(a^2\) even \(\Rightarrow a\) even | the parity step inside the \(\sqrt2\) proof | No — derive it |
| \(N = p_1 p_2 \cdots p_n + 1\) | the new number in the proof that there are infinitely many primes | No — learn it |
Check your understanding
1. Prove by contradiction that there is no greatest multiple of 7.
Assume there is a greatest multiple of 7, \(M = 7k\) for some integer \(k\). Then \(M + 7 = 7(k+1)\) is also a multiple of 7, and \(M + 7 > M\). This contradicts \(M\) being the greatest, so there is no greatest multiple of 7.
2. Prove by contradiction that if \(n^2\) is odd, then \(n\) is odd.
Assume \(n^2\) is odd and \(n\) is even, so \(n = 2k\). Then \(n^2 = 4k^2 = 2(2k^2)\), which is even: a contradiction. So if \(n^2\) is odd, \(n\) is odd.
3. Express \(\frac{5}{(x-1)(x+4)}\) in partial fractions.
\(5 \equiv A(x+4) + B(x-1)\). \(x = 1\): \(5 = 5A\), \(A = 1\); \(x = -4\): \(5 = -5B\), \(B = -1\). So \(\frac{1}{x-1} - \frac{1}{x+4}\).
4. Write down the form of the partial fractions of \(\frac{2x+3}{x^2(x-5)}\), then find the constants.
\(x^2\) is a repeated factor (\(x\) twice), so the form is \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-5}\). From \(2x + 3 \equiv Ax(x-5) + B(x-5) + Cx^2\): \(x = 0\) gives \(B = -\frac35\); \(x = 5\) gives \(13 = 25C\), \(C = \frac{13}{25}\); the \(x^2\) terms give \(A + C = 0\), so \(A = -\frac{13}{25}\).
5. Why must \(\frac{x^2+4}{(x+1)(x-3)}\) be divided before it is split, and what is the result?
The numerator and denominator both have degree 2, so the fraction is improper. \(x^2 + 4 = (x^2 - 2x - 3) + (2x + 7)\), giving \(1 + \frac{2x+7}{(x+1)(x-3)}\). Then \(2x + 7 \equiv A(x-3) + B(x+1)\): \(x = -1\) gives \(A = -\frac54\), \(x = 3\) gives \(B = \frac{13}{4}\). Result: \(1 - \frac{5}{4(x+1)} + \frac{13}{4(x-3)}\).
6. Prove by contradiction that \(\sqrt2 - 5\) is irrational.
Assume \(\sqrt2 - 5 = \frac{r}{s}\) with \(r, s\) integers, \(s \ne 0\). Then \(\sqrt2 = \frac{r}{s} + 5 = \frac{r + 5s}{s}\), a ratio of integers, so \(\sqrt2\) would be rational. This contradicts worked example 1, so \(\sqrt2 - 5\) is irrational.
Specification checklist
- (P4 1.1) I can set out a proof by contradiction: state the negation of the statement precisely, reason correctly from it to an impossibility, and write the conclusion.
- (P4 1.1) I can prove that \(\sqrt2\) is irrational, including the step that if \(a^2\) is even then \(a\) is even, and adapt the proof to roots such as \(\sqrt3\).
- (P4 1.1) I can prove that there are infinitely many prime numbers.
- (P4 1.1) I can use contradiction for unfamiliar statements, such as results about rational and irrational numbers, equations with no integer solutions and "no greatest" results.
- (P4 2.1) I can split a proper fraction whose denominator has distinct linear factors (two or three) into partial fractions, by substitution or by comparing coefficients.
- (P4 2.1) I can find partial fractions when the denominator has a repeated linear factor.
- (P4 2.1) I can recognise an improper fraction, divide first, and then split the remaining proper fraction.